已知關於x的方程x²;-3x+k=0的兩根立方和為0,則k= 我這樣解哪個地方出錯了?請您指點一下 依題意可知:x1三次方+x2三次方=0 所以有兩種結論:①x1=x2=0②x1+x2=0,也就是互為相反數 檢驗②:由韋達定理知-b/a=3≠0,所以不成立 那麼只有是①了, 所以兩根相等 b²;-4ac=0 ∴9-4k=0 ∴k=9/4

已知關於x的方程x²;-3x+k=0的兩根立方和為0,則k= 我這樣解哪個地方出錯了?請您指點一下 依題意可知:x1三次方+x2三次方=0 所以有兩種結論:①x1=x2=0②x1+x2=0,也就是互為相反數 檢驗②:由韋達定理知-b/a=3≠0,所以不成立 那麼只有是①了, 所以兩根相等 b²;-4ac=0 ∴9-4k=0 ∴k=9/4


x1^3+x2^3=0推不出你所說的兩個結論,因為這裡可能為虛根.應該如此:
x1^3+x2^3=(x1+x2)(x1^2+x1x2+x2^2)=(x1+x2)[(x1+x2)^2-x1x2]=3[3^2-k]=0--> k=9



已知關於x的方程:x的平方+(2k+1)x+k的平方-2=0的兩實根的平方和比兩根積的3倍少10,求k的值


x^2+(2k+1)x+k^2-2=0
x1+x2=-2k-1 x1x2=k^2-2
x1^2+x2^2=3x1x2-10
(x1+x2)^2=5x1x2-10
(2k+1)^2=5k^2-20
4k^2+4k+1=5k^2-20
k^2-4k-21=0
(k-7)(k+3)=0
所以k=7或k=-3



若a是方程2x²;-x-3=0的一個解,則6a-3a的值是?


 ;



已知關於x的方程x²;+(a-1)x+1=0的兩個實數根為α,β.求a的取值範圍?


(a-1)²;-4≥0
(a-1+2)(a-1-2)≥0
(a+1)(a-3)≥0
∴a≤-1或a≥3



求方程x²;-mx+2m-1=0兩根平方和的取值範圍


首先b²;-4ac≥0,即m²;-4(2m-1)≥0,解得m≥4+2√3或m≤4-2√3
根據韋達定理x1+x2=m.x1*x2= 2m-1
(x1)²;+(x2)²;=(x1+x2)²;-2x1x2
= m²;- 4m+2
=(m-2)²;-2
當m= 4-2√3時,有最小值14-8√3
範圍是[14-8√3,+∞)



關於x的方程mx2+(2m+1)x+m=0有兩個不相等的實數根,則m的取值範圍是()
A. m>-14B. m<-14C. m>4D. m>-14且m≠0


∵關於x的方程mx2+(2m+1)x+m=0有兩個不相等的實數根,∴方程為一元二次方程,∴△=(2m+1)2-4m•m>0且m≠0,∴4m2+1+4m-4m2>0,∴4m>-1,∴m>-14且m≠0.故選D.



若方程x²;+(m+2)x+5=0只有負根,則m的取值範圍
若方程x²;+(m+2)x+5=0只有負根,則m的取值範圍?


x1x2=5>0
有兩個負根.
△=(m+2)^2-20≥0且x1+x2=-(m+2)



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